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Damping an oscillatory motion: friction on an object connected to a spring

Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in [link] , but there is simple friction between the object and the surface, and the coefficient of friction μ k size 12{μ rSub { size 8{k} } } {} is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at v = 0 size 12{v=0} {} ? The force constant of the spring is k = 50 . 0 N/m size 12{k="50" "." 0`"N/m"} {} .

 The given figure (a) shows a spring on a frictionless surface attached to a bar or wall from the left side and on the right side of the spring, there is an object attached with mass m. Its amplitude is given by X, and X is equal to zero at the equilibrium level. Force F is applied to it from the right side, represented by a red arrow pointing toward the left and velocity v is equal to zero. An arrow showing the direction of force is also given alongside this figure as well as with the other four figures. The energy of the object is half k x squared.           In the given figure (b), after force is applied, the object moves to the left, compressing the spring slightly. The displacement of the object from its initial position is indicated by dots. The force F, here is equal to zero and velocity v, is maximum in the negative direction or the left. The energy of the object in this case is half m times negative v-max whole squared.           In the given figure (c), the spring has been compressed the maximum limit, and the amplitude is minus X. Now the force is toward the right, indicated here with a red arrow pointing to the right and the velocity, v, is zero. The energy of the object now is half k times negative x whole squared.           In the given figure (d), the spring is shown released from its compressed position and the object has moved toward the right side to reach the equilibrium level. Here, F is equal to zero, and the velocity, v, is the maximum. The energy of the object becomes half k times v max squared.           In the given figure (e), the spring has been stretched loose to the maximum possible limit and the object has moved to the far right. Now the velocity v, here is equal to zero and the direction of force is toward the left. As shown here, F is equal to zero. The energy of the object in this case is half k times x squared.
The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.

Strategy

This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems.

Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question.

Solution a

  1. Choose the proper equation: Friction is f = μ k mg size 12{F=μ rSub { size 8{k} } ital "mg"} {} .
  2. Identify the known values.
  3. Enter the known values into the equation:
    f = (0.0800) (0 .200 kg) (9 .80 m / s 2 ) . size 12{f=0 "." "0800" times 0 "." "200"`"kg" times 9 "." 8`"ms" rSup { size 8{"-2"} } } {}
  4. Calculate and convert units: f = 0.157 N . size 12{F=μ rSub { size 8{k} } ital "mg"} {}

Discussion a

The force here is small because the system and the coefficients are small.

Solution b

Identify the known:

  • The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down.
  • Energy is not conserved as the mass oscillates because friction is a non-conservative force.
  • The motion is horizontal, so gravitational potential energy does not need to be considered.
  • Because the motion starts from rest, the energy in the system is initially PE el,i = ( 1 / 2 ) kX 2 size 12{ ital "PE" rSub { size 8{e1} } = \( 1/2 \) ital "kX" rSup { size 8{2} } } {} . This energy is removed by work done by friction W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "fd"} {} , where d size 12{x} {} is the total distance traveled and f = μ k mg size 12{f=μk ital "mg"} {} is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so PE e1,f = ( 1 / 2 ) kx 2 size 12{"PE" rSub { size 8{"e1,f"} } = \( 1/2 \) ital "kx" rSup { size 8{2} } } {} where x size 12{x} {} is the final position and is given by
    F el = f kx = μ k mg x = μ k mg k . alignl { stack { size 12{F rSub { size 8{"el"} } =f} {} #ital "kx"=μ rSub { size 8{k} } ital "mg" {} # x= { {μ rSub { size 8{k} } ital "mg"} over {k} } {}} } {}
  1. By equating the work done to the energy removed, solve for the distance d size 12{x} {} .
  2. The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use:
    W nc = Δ KE + PE = PE el,f PE el,i = 1 2 k μ k mg k 2 X 2 . size 12{W rSub { size 8{"nc"} } =Δ left ("KE"+"PE" right )="PE" rSub { size 8{"el,f"} } - "PE" rSub { size 8{"el,i"} } = { {1} over {2} } k left ( left ( { {μ rSub { size 8{k} } ital "mg"} over {k} } right ) rSup { size 8{2} } - X rSup { size 8{2} } right )} {}
  3. Recall that W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "Fd"} {} .
  4. Enter the friction as f = μ k mg size 12{F=μ rSub { size 8{k} } ital "mg"} {} into W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "Fd"} {} , thus
    W nc = μ k mgd . size 12{W size 8{"nc"}=μ rSub { size 8{k} } ital "mgd"} {}
  5. Combine these two equations to find
    1 2 k μ k mg k 2 X 2 = μ k mgd . size 12{ { {1} over {2} } k left ( left ( { {μ rSub { size 8{k} } } over {k} } right ) rSup { size 8{2} } - X rSup { size 8{2} } right )= - μ rSub { size 8{k} } } {}
  6. Solve the equation for d size 12{x} {} :
    d = k 2 μ k mg ( X 2 ( μ k mg k ) 2 ) . size 12{d= { { { {1} over {2} } ital "kX" rSup { size 8{2} } } over {μ rSub { size 8{k} } } } } {}
  7. Enter the known values into the resulting equation:
    d = 50 . 0 N/m 2 0 . 0800 0 . 200 kg 9 . 80 m/s 2 0 . 100 m 2 0 . 0800 0 . 200 kg 9 . 80 m/s 2 50 . 0 N/m 2 . size 12{d= { {"50" "." 0" N/m"} over {2 left (0 "." "0800" right ) left (0 "." "200"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} } left ( left (0 "." "100"" m" right ) rSup { size 8{2} } - left ( { { left (0 "." "0800" right ) left (0 "." "200"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} over {"50" "." 0" N/m"} } right ) rSup { size 8{2} } right )} {}
  8. Calculate d size 12{x} {} and convert units:
    d = 1 . 59 m . size 12{d=1 "." "59"`m} {}

Discussion b

This is the total distance traveled back and forth across x = 0 size 12{x=0} {} , which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than d / X = ( 1 . 59 m ) / ( 0 . 100 m ) = 15 . 9 size 12{d/X= \( 1 "." "59"`m \) / \( 0 "." "100"`m \) ="15" "." 9} {} because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to x = 0 size 12{x=0} {} for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position x = 0 size 12{x=0} {} a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position.

This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life.

Why are completely undamped harmonic oscillators so rare?

Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator.

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Describe the difference between overdamping, underdamping, and critical damping.

An overdamped system moves slowly toward equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium.

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Section summary

  • Damped harmonic oscillators have non-conservative forces that dissipate their energy.
  • Critical damping returns the system to equilibrium as fast as possible without overshooting.
  • An underdamped system will oscillate through the equilibrium position.
  • An overdamped system moves more slowly toward equilibrium than one that is critically damped.

Conceptual questions

Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)

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How would a car bounce after a bump under each of these conditions?

  • overdamping
  • underdamping
  • critical damping

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Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics?

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Problems&Exercises

The amplitude of a lightly damped oscillator decreases by 3 . 0% size 12{3 "." 0%} {} during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

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Questions & Answers

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Ex. On an inclined plane, gravitational potential energy, friction energy/work and spring potential energy. (Let's say that the spring is keeping the box from sliding down the slope.) How do we use this in the equation? I'm so confused
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Oh! And if there's kinetic energy that is exerting a force opposite to the spring, what do we do?
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Then we subtract the k.e. from force exerted from newton's 2nd law.
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50 N...(50 *1.732)N
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The abacus (plural abaci or abacuses), also called a counting frame, is a calculating tool that was in use in Europe, China and Russia, centuries before the adoption of the written Hindu–Arabic numeral system
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a load of 20N on a wire of cross sectional area 8×10^-7m produces an extension of 10.4m. calculate the young modules of the material of the wire is of length 5m
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Young's modulus = stress/strain strain = extension/length (x/l) stress = force/area (F/A) stress/strain is F l/A x
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Practice Key Terms 3

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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